F.2 Maths

let X be the hundred digit and Y be the ten digit, we have.
(100Y+10Y+X)-(100X+10Y+Y)=198-------------------(1)
&
X+Y=6---------------------------------------(2)
from (2) we haveY=6-X-------------------------------------------(3)
substitute (3) into (1), we have
(100(6-X)+10(6-X)+X)-(100X+10(6-X)+(6-X))=198
600-100X+60-10X+X-100X-60+10X-6+X=198
-198X=198-600-60+60+6
-198X=-396
X=2

let a be the united digit and b be the hundred digit
then a+b=6 .....(1) ( sum of the hundreds and tens digit is 6)
a+10a+100b+198=b+10a+100a
(new three-digit number is greater than the original number by 198)
99b+198=99a
b+2=a......(2)
subs. 2 into 1 : b+2+b=6
b=2
from1, a=4
so the original number is 244.